What are the solutions to the equation -x^2-5x=x^2-3x-4 ?

The solutions to the equation -x^2-5x=x^2-3x-4 are x=1,x=-2

Find the zeros of -x^2-5x=x^2-3x-4

The zeros of -x^2-5x=x^2-3x-4 are x=1,x=-2

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- x^{2} - 5 x = x^{2} - 3 x - 4

x = 1, x = - 2

Solution steps

- x^{2} - 5 x = x^{2} - 3 x - 4

Switch sides

x^{2} - 3 x - 4 = - x^{2} - 5 x

Move

5 x

to the left side

x^{2} + 2 x - 4 = - x^{2}

Move

x^{2}

to the left side

2 x^{2} + 2 x - 4 = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 2 \pm 2 ^{2} - 4 \cdot 2 ( - 4 )}{2 \cdot 2}

2^{2} - 4 \cdot 2 (- 4) = 6

x_{1, 2} = \frac{- 2 \pm 6}{2 \cdot 2}

Separate the solutions

x_{1} = \frac{- 2 + 6}{2 \cdot 2}, x_{2} = \frac{- 2 - 6}{2 \cdot 2}

x = \frac{- 2 + 6}{2 \cdot 2} : 1

x = \frac{- 2 - 6}{2 \cdot 2} : - 2

The solutions to the quadratic equation are:

x = 1, x = - 2

\frac{( 1 - 1 ) ^{0}}{1 ^{0}}

\frac{x}{5} + 3 > 2

25 a^{2} - 100

\frac{6 i}{1 + 5 i}

x^{2} + 2 x - 168

What are the solutions to the equation -x^2-5x=x^2-3x-4 ?
The solutions to the equation -x^2-5x=x^2-3x-4 are x=1,x=-2
Find the zeros of -x^2-5x=x^2-3x-4
The zeros of -x^2-5x=x^2-3x-4 are x=1,x=-2